The value of k , for which the function

$$\mathrm{f}(x)= \begin{cases}\left(\frac{4}{5}\right)^{\frac{\ln 4 x}{\tan 5 x}}, & 0< x< \frac{\pi}{2} \\ \mathrm{k}+\frac{2}{5} & , x=\frac{\pi}{2}\end{cases}$$

is continuous at $x=\frac{\pi}{2}$, is

$\lim _\limits{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ is equal to

The value of $\lim _\limits{x \rightarrow 0} \frac{x}{|x|+x^2}$ is

If $\mathrm{f}(x)=\frac{1+\cos \pi x}{\pi(1-x)^2}$, for $x \neq 1$ is continuous at $x=1$, then $\mathrm{f}(1)$ is equal to